Question

The first term of an arithmetic sequence is 7 and the common difference is 4. Find the 30th term of the sequence.

a1 =
n =
d =
x =
y =

Answer 1

Answer:

$given \\ fist \: term(a) = 7 \\ common \: difference(d) = 4 \\ we \: know \: that \: nth \: term \: = a + d(n - 1) \\ therefore \: 30th \: term = 7 + 4(30 - 1) \\ = 11 \times 29 \\ = 319$

Hope this helps you mate, wish you all the best.

Answer 2

Solution:
Let a be the first term and d be the common difference respectively.
:. tn=a+(n-1)d…….formula
:.here, a=7, d=4, n=30
:. t30=7+(30-1)(4)
:. t30=7+(29)(4)
:. t30=7+116
:. t30=123….(Ans)

The 30th term is 123…(Ans)